https://doi.org/10.1007/s00574-020-00219-4 CORRECTION
Correction to: A Family of Foliations with One Singularity
S. C. Coutinho1 ·Filipe Ramos Ferreira1
© Sociedade Brasileira de Matemática 2020
Abstract
We fix a mistake in the argument leading to the proof that the family of foliations introduced in the paper does not have an algebraic solution apart from the line at infinity
Correction to: Bull Braz Math Soc, New Series
https://doi.org/10.1007/s00574-019-00183-8
We use throughout the notation of section 5 of our paper. A mistake was introduced in the paper by our assumption in Proposition 5.1 thatφm =1, because it requires that we divide all the coefficients of the algebraic solution f byφm. However, in doing that we get a polynomial whose coefficients do not belong toR, which precludes the divisibility arguments used to get contradictions in Propositions 5.1 and 5.3 and in Theorem 5.5. We show here that it is possible to arrive at the same contradictions without requiring f to be inR[y,z].
We will assume that
A=αyk−3, B =βyk−1 and q =ρy2, (1) since this is the only case needed for Theorem 5.5 and it simplifies the equations. We will also assume thatk ≥ 7, to avoid a multiplicity of cases. Examples of smaller degree can be handled using a computer algebra system.
The original article can be found online athttps://doi.org/10.1007/s00574-019-00183-8.
B
S. C. Coutinho collier@dcc.ufrj.br Filipe Ramos Ferreira frferreira9@gmail.com1 Departamento de Ciência da Computação, Instituto de Matemática, Universidade Federal do Rio de Janeiro, P.O. Box 68530, 21945-970 Rio de Janeiro, RJ, Brazil
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Bull Braz Math Soc, New Series (2021) 52:767–769
Published online: 6 July 2020
S. C. Coutinho, F. R. Ferreira
Proof of Proposition 5.1 The only point in the proof where a divisibility argument is used is at the very end of the proof, after the equation
zk−2φm+2−2Aφm+2= −(d−m)h. (2) Note that there was a typo on the first term of (2). Since we have already shown that em+2=m−2, it follows that
deg(φm+2)=(m+2)−(m−2)=4.
Letφm+2=e0y4+e1y3z+e2y2z2+e3yz3+e4z4. The terms−2αeiyk+1−izi on the left hand side of (2) have no correspondent on the right side fori =1, . . . ,4 and the same holds for 4e0y3zk−2. Thus,φm+2=0, which gives the desired contradiction
becaused =mandh=0 by hypothesis.
Proof of Proposition 5.3 Divisibility arguments are used twice in the proof of this proposition. The first time is right at the beginning, where we assume thatμ(m)= ε0(m)=ε2(m)=m. However, in this case, the equation that results from (5.7) is (2) and we have already shown that it leads to a contradiction. The second place where a divisibility condition intervenes is at the end of the proof. Letm<k≤dbe the small- est integer such thatek =0. We know such an integer exists because, by Corollary 3.2,ed=0. But, having proved thatej =3m−2j, we have that
ek−1=3m−2(k−1)=(3m−2k)+2=ek+2=2;
so thatε0(k−1)=2,ε1(k−1)=1 andε1(k−1)+σ =0. Thus,μ(k−1)=0.
Since, eitherε2(k−1) = ek+1+2 ≥ 2 orψk+1 = 0, it follows from (5.7), with r =k−1, thatθ3(k−1) = −αψm+1ψk =0, which implies, by Lemma 5.1, that ψk=0, contradicting the formula forψk in (5.8).
Proof of Theorem 5.5 It follows from the hypothesis of the theorem that the proposi- tions in section 5 can be applied toωu0. Moreover, by Theorem 2.1, we can assume that the coefficients of f belong to Q(α, β, ρ,c). It follows from Proposition 5.1 that em = m and from Proposition 5.3 that em+1 = m−1. Hence, σ = 0 and ε0(m)= ε1(m) =m. Ifμ(m)= ε2(m) < mthen we can deduce from (5.9) with r =mthatθ2(m)= 2αyk−3ψm+2 =0. Hence,ψm+2 = 0, which is only possible whenφm+2 =0. However, if this were the case, we would haveμ(m)=m, so that again by (5.9) withr =m,
−(d−m)η+2βψm+1−αψm2+1=0,
contradicting our assumption thatψm+1∈ Q(α, β, ρ,c). Sinceε2(m) >mimplies that μ(m) = m, we also get a contradiction in this case. Hence, ε2(m) = m and em+2=m−2. It follows fromem+1=m−1 andem+2=m−2 that deg(φm+1)=2
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Correction to: A Family of Foliations with One Singularity
and deg(φm+2)=4; so we can set φm+1=
2
j=0
sjyjz2−j and φm+2=
4
i=0
eiyiz4−i,
whereei,sj ∈Q(α, β, ρ,c)for 0≤ j ≤2 and 0≤i ≤4. Now, equation (5.6) with r=mgives
(m−d)h+yk−3
βy2φm+1+2αφm+2−αφm2+1
=zk−2
ρy2φm+1+φm+2−φm+1φm+1
.
Equating the coefficients ofyk−2z3andzk+1on both sides of the previous equation, we get
2e1α−2s0s1α=0 and c(m−d)=e1−s0s1.
Since α = 0 the first equation givese1 = s0s1. Substituting this into the second equation above we getc(m−d)= 0, which is a contradiction becausec= 0 and
m<d.
Unfortunately, the specialization from genericu tou0 =(αyk−3, βyk−1, ρy2,c) in the proof of Corollary 5.6 requires that Propositions 5.1 and 5.3 hold in the generic case. Thus, the statement of this corollary should now be taken as a conjecture.
Acknowledgements During the preparation of the paper the first author was partially supported by CNPq Grant 304543/2017-9 and the second author by a Grant PIBIC(CNPq). We also benefited from the access to on-line journals provided by CAPES.
Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
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